Convert Sorted Array to Binary Search Tree

Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree.

Example 1:
Input: nums = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: [0,-10,5,null,-3,null,9] is also accepted:

Example 2:
Input: nums = [1,3]
Output: [3,1]
Explanation: [1,null,3] and [3,1] are both height-balanced BSTs.
Constraints:

1 <= nums.length <= 104
-104 <= nums[i] <= 104
nums is sorted in a strictly increasing order.
 

思路一：要構建二叉排序樹，考慮使用遞迴的思想，每次取一箇中值作為節點的值，左邊的所有值用於構造節點的左子樹，右邊的所有值用於構造節點的右子樹。

由於樹是排序的，左節點的值一定要小於右節點，所以判斷中值的時候注意取整條件

public TreeNode sortedArrayToBST(int[] nums) {
    return sortedArrayToBST(nums, 0, nums.length - 1);
}

public TreeNode sortedArrayToBST(int[] nums, int begin, int end) {
    int mid = (begin + end) % 2 == 0 ? (begin + end) / 2 : (begin + end) / 2 + 1;
    TreeNode node = new TreeNode(nums[mid]);

    if (begin < mid) {
        node.left = sortedArrayToBST(nums, begin, mid -  1);
    }

    if (mid < end) {
        node.right = sortedArrayToBST(nums, mid + 1, end);
    }

    return node;
}